Do Problem 6.16 on page 175 in the textbook.
EE 306, Fall 2009
Problem Set 6 Solutions
Due:
Not to be turned in
Yale N. Patt, Instructor
TAs: Aater Suleman, Chang Joo Lee, Ameya Chaudhari, Antonius Keddis, Arvind Chandrababu, Bhargavi Narayanasetty, Eshar Ben-dor, Faruk Guvenilir, Marc Kellermann, RJ Harden
Note: This problem set is unusually long, and
is not to be turned in. We have put it together and handed it out to
give you some challenging examples to help you prepare for the final
exam.
(Adapted from 10.9) The input stream of a stack is a list of
all the elements we pushed onto the stack, in the order that we
pushed them. The input stream from Excercise 10.8 from page 284 of
the book for example is ABCDEFGHIJKLM
The output stream is a list
of all the elements that are popped off the stack in the order that
they are popped off.
a. If the input stream is ZYXWVUTSR, create
a sequence of pushes and pops such that the output stream is
YXVUWZSRT.
Push Z
Push Y
Pop Y
Push
X
Pop X
Push W
Push V
Pop V
Push U
Pop U
Pop
W
Pop Z
Push T
Push S
Pop S
Push R
Pop R
Pop
T
b. If the input stream is ZYXW, how many different
output streams can be created.
14 different
output streams.
Do Problem 6.16 on page 175 in the textbook.
(Adapted from 10.6) Rewrite the PUSH and POP routines such
that the stack on which they operate holds elements that take up two
memory locations each.
The problem assumes
that each element of the value being pushed on the stack is
32-bits.
For the PUSH, assume bits [15:0] of that value to be
pushed are in R0 and bits [31:16] are in R1.
For the POP, bits
[15:0] will be popped into R0 and bits [31:16] will be popped into
R1.
Also assume the lower order bits of the number being pused or
popped are located in the smaller address in memroy. For example if
the two memory locations to be used to store the number are x2FFF
and x2FFE, bits [15:0] will be stored in x2FFE and [31:16] will be
stored in x2FFF.
PUSH:
ADD R6, R6, #-2
STR R0, R6,
#0
STR R1, R6, #1
POP:
LDR R0, R6, #0
LDR R1, R6,
#1
ADD R6, R6, #2
Consider the following LC-3 assembly language program. Assumming that the memory locations DATA get filled before the program executes, what is the relationship between the final values at DATA and the initial values at DATA?
.ORIG x3000
LEA R0, DATA
AND R1, R1, #0
ADD R1, R1, #9
LOOP1 ADD R2, R0, #0
ADD R3, R1, #0
LOOP2 JSR SUB1
ADD R4, R4, #0
BRzp LABEL
JSR SUB2
LABEL ADD R2, R2, #1
ADD R3, R3, #-1
BRp LOOP2
ADD R1, R1, #-1
BRp LOOP1
HALT
DATA .BLKW #10
SUB1 LDR R5, R2, #0
NOT R5, R5
ADD R5, R5, #1
LDR R6, R2, #1
ADD R4, R5, R6
RET
SUB2 LDR R4, R2, #0
LDR R5, R2, #1
STR R4, R2, #1
STR R5, R2, #0
RET
.ENDThe final values at DATA will be sorted in ascending order.
During the initiation of the interrupt service routine, the
N, Z, and P condition codes are saved on the stack. By means of a
simple example show how incorrect results would be generated if the
condition codes were not saved. Also, clearly describe the steps
required for properly handling an interrupt.
Lets
take the following program which adds 10 numbers starting at memory
location x4000 and stores the result at x5000.
.ORIG x3000
LD R1, PTR
AND R0, R0, #0
LD R2, COUNT
LOOP LDR R3, R1, #0
ADD R0, R0, R3
ADD R1, R1, #1
ADD R2, R2, #-1
BRp LOOP
STI R0, RESULT
HALT
PTR .FILL x4000
RESULT .FILL x5000
COUNT .FILL #10
If the condition codes were not saved as part
of initiation of the interrupt service routine, we could end up with
incorrect results. In this program, take the case when an interrupt
occurred during the processing of the instruction at location x3006
and the condition codes were not saved. Let R2 = 5 and hence the
condition codes would be N=0, Z=0, P=1, before servicing the
interrupt. When control is returned to the instruction at location
x3007, the BRp instruction, the condition codes depend on the
processing within the interrupt service routine. If they are N=0,
Z=1, P=0, then the BRp is not taken. This means that the result
stored is just the sum of the first five values and not all
ten.
Steps for handling interrupts:
Saving the State of the machine
Loading the state of the interrupt
Service the Interrupt
RTI
Note: In-depth explanation of interrupt
handling on pages 259-261 of the texbook.
The program below counts the number of zeros in a 16-bit word. Fill in the missing blanks below to make it work.
.ORIG x3000
AND R0, R0, #0
LD R1, SIXTEEN
LD R2, WORD
A BRn B
ADD R0, R0, #1
B ADD R1, R1, #-1
BRz C
ADD R2, R2, R2
BR A
C ST R0, RESULT
HALT
SIXTEEN .FILL #16
WORD .BLKW #1
RESULT .BLKW #1
.END
After you have the correct answer above, what one instruction
can you change (without adding any instructions) that will make the
program count the number of ones instead?
Replace
the BRn instruction with a BRzp.
Fill in the missing blanks so that the subroutine below
implements a stack multiply. That is it pops the top two elements
off the stack, multiplies them, and pushes the result back on the
stack. You can assume that the two numbers will be non-negative
integers (greater than or equal to zero) and that their product will
not produce an overflow. Also assume that the stack has been
properly initialized, the PUSH and POP subroutines have been written
for you and work just as described in class, and that the stack will
not overflow or underflow.
Note: All blanks must be filled
for the program to operate correctly.
MUL ST R7, SAVER7
ST R0, SAVER0
ST R1, SAVER1
ST R2, SAVER2
ST R5, SAVER5
AND R2, R2, #0
JSR POP
ADD R1, R0, #0
JSR POP
ADD R1, R1, #0
BRz DONE
AGAIN ADD R2, R2, R0
ADD R1, R1, #-1
BRp AGAIN
DONE ADD R0, R2, #0
JSR PUSH
LD R7, SAVER7
LD R0, SAVER0
LD R1, SAVER1
LD R2, SAVER2
LD R5, SAVER5
RET
The program below calculates the closest integer greater than
or equal to the square root of the number stored in NUM, and prints
it to the screen. That is, if the number stored in NUM is 25, "5"
will be printed to the screen. If the number stored in NUM is 26,
"6" will be printed to the screen. Fill in the blanks
below to make the program work.
Note: Assume that the value
stored at NUM will be between 0 an 81.
.ORIG x3000
AND R2, R2, #0
LD R3, NUM
BRz OUTPUT
NOT R3, R3
ADD R3, R3, #1
OUTLOOP ADD R2, R2, #1
ADD R0, R2, #0
AND R1, R1, #0
INLOOP ADD R1, R1, R2
ADD R0, R0, #-1
BRp INLOOP
ADD R1, R1, R3
BRn OUTLOOP
OUTPUT LD R0, ZERO
ADD R0, R0, R2
TRAP x21
HALT
NUM .BLKW 1
ZERO .FILL x30
.END
The figure below shows the part of the LC-3 data path that
deals with memory and I/O. Note the signals labeled A through F. A
is the memory enable signal, if it is 1 memory is enabled, if it is
0, memory is disabled. B, C, and D are the load enable signals for
the Device Registers. If the load enable signal is 1, the register
is loaded with a value, otherwise it is not. E is the 16-bit output
of INMUX, and F is the 2-bit select line for INMUX.
The
initial values of some of the processor registers and the I/O
registers, and some memory locations are as follows:
|
R0 = x0000 |
KBSR = x8000 |
M[x3009] = xFE00 |
During the entire instruction cycle, memory is accessed
between one and three times (why?). The following table lists two
consecutive instructions to be executed on the LC-3. Complete the
table with the values that each signal or register takes right after
each of the memory accesses performed by the instruction. If an
instruction does not require three memory accesses, draw a line
accross the unused accesses. To help you get started, we have filled
some of the values for you.
|
PC |
Instruction |
Access |
MAR |
A |
B |
C |
D |
E[15:0] |
F[1] |
F[0] |
MDR |
|
x3000 |
LD R0, x9 |
1 |
x3000 |
1 |
0 |
0 |
0 |
x2009 |
1 |
1 |
x2009 |
|
2 |
x300A |
1 |
0 |
0 |
0 |
xFE02 |
1 |
1 |
xFE02 |
||
|
3 |
--------- |
--- |
--- |
--- |
--- |
--------- |
------ |
------ |
--------- |
||
|
x3001 |
LDR R0, R0, #0 |
1 |
x3001 |
1 |
0 |
0 |
0 |
x6000 |
1 |
1 |
x6000 |
|
2 |
xFE02 |
0 |
0 |
0 |
0 |
x0061 |
0 |
0 |
x0061 |
||
|
3 |
--------- |
--- |
--- |
--- |
--- |
--------- |
------ |
------ |
--------- |
Note: This problem is NOT easy. In fact, it took me a while
to solve it, and I am supposed to be an expert on 306 material. So,
if you are struggling to pass this course, I suggest you ignore it.
On the other hand, if you are a hot shot and think no problem is
beyond you, then by all means go for it. We put it on the problem
set to keep some of the hot shots out of mischief. We would not put
it on the final, because we think it is too difficult to put on the
exam.
A programmer wrote this program to do something useful.
He, however, forgot to comment his code, and now can't remember what
the program is supposed to do. Your job is to save him the trouble
and figure it out for him. In 20 words or fewer tell us what
valuable information the program below provides about the value
stored in memory location INPUT. Assume that there is a non-zero
value at location INPUT before the program is executed.
HINT:
When testing different values of INPUT pay attention to their bit
patterns. How does the bit pattern correspond to the RESULT?
.ORIG x3000
LD R0, INPUT
AND R3, R3, #0
LEA R6, MASKS
LD R1, COUNT
LOOP LDR R2, R6, #0
ADD R3, R3, R3
AND R5, R0, R2
BRz SKIP
ADD R3, R3, #1
ADD R0, R5, #0
SKIP ADD R6, R6, #1
ADD R1, R1, #-1
BRp LOOP
ST R3, RESULT
HALT
COUNT .FILL #4
MASKS .FILL 0xFF00
.FILL 0xF0F0
.FILL 0xCCCC
.FILL 0xAAAA
INPUT .BLKW 1
RESULT .BLKW 1
.END
This program identifies the most significant bit position that is set in the value stored at INPUT and stores that bit position in RESULT. For example, if INPUT contained the value 0010 0100 0101 0110, RESULT would contain the value 13 since bit 13 is the most significant bit postition that is a 1.
Figure out what the following program does.
.ORIG X3000
LEA R2, C
LDR R1, R2, #0
LDI R6, C
LDR R5, R1, #-3
ST R5, C
LDR R5, R1, #-4
LDR R0, R2, #1
JSRR R5
AND R3, R3, #0
ADD R3, R3, #7
LEA R4, B
A STR R4, R1, #0
ADD R4, R4, #2
ADD R1, R1, #1
ADD R3, R3, #-1
BRP A
HALT
B ADD R2, R2, #1
LDR R0, R2, #0
JSRR R5
TRAP X29
ADD R2, R2, #15
ADD R0, R2, #3
LD R5, C
TRAP X2B
ADD R2, R2, #5
LDR R0, R2, #0
JSRR R5
TRAP X27
JSRR R5
JSRR R6
C .FILL X25
.STRINGZ "EE306 and tests are awesome"
.END
The short answer is that the program outputs “EE some” this is because we over write the trap vector table. Below is a commented version of the program to help you see what is going on.
.ORIG X3000 LEA R2, C LDR R1, R2, #0 ; load x25 into R1 LDI R6, C ; loads the starting address of the HALT trap service routine into R6 LDR R5, R1, #-3 ; loads the starting address of (x25 – 3) trap x22 (puts) into R5 ST R5, C ; stores the starting address or puts into C LDR R5, R1, #-4 ; loads the starting address of (x25 – 4) trap x21 (out) into r5 LDR R0, R2, #1 ; loads R0 with the first charater of the stringz “E” JSRR R5 ; does the out routine (outputs “E” to the display) AND R3, R3, #0 ; clears r3 ADD R3, R3, #7 ; makes r3 7 LEA R4, B ; loads the address of B into r4 ;NOTE Loop A overwrites the trap vector table, x25 to x2b ; This makes trap x25 – trap x2b point to this program, see label B and below A STR R4, R1, #0 ; overwrites the trap vector with the address in R4 ADD R4, R4, #2 ADD R1, R1, #1 ADD R3, R3, #-1 BRP A HALT ; What does this do? Trap x25, what is now at memory location x25? ;In the following section <- trap xY indicates what address is in memory location Y B ADD R2, R2, #1 ; <- trap x25 (makes R2 point to the first character in the stringz “E”) LDR R0, R2, #0 ; (loads r0 with the ascci code for “E“) JSRR R5 ; <- trap x26 (what is in r5? The starting address of out,outputs “E” on the screen) TRAP X29 ADD R2, R2, #15; <- trap x27 (makes r2 points to the (6 + 15) 21th character of the .stringz ADD R0, R2, #3 ; (makes to point to the (21+3) 24th character of the stringz the s in awesome) LD R5, C ; <- trap x28 (LD R5, C loads r5 with the starting address of puts) TRAP X2B ADD R2, R2, #5 ; <- trap x29 (“makes R2 point to the 6th character in the .stringz “ “) LDR R0, R2, #0 ; (loads r0 with the ascci code for “ “) JSRR R5 ; <- trap x2a (outputs a space on the screen) TRAP X27 JSRR R5 ; <- trap x2b (jsrr to puts outputs “some” to the screen) JSRR R6 ; remember r6 contains the starting address of trap x25 (halt) so this halts C .FILL X25 .STRINGZ "EE306 and tests are awesome" .END